Question

a hot air balloon is filled with ( 1.33\times10^{6}) l of an ideal gas on a cool morning (( 11^{circ}c)). the air is heated to ( 121^{circ}c). what is the volume of the air in the balloon after it is heated? assume that none of the gas escapes from the balloon.

Explanation:

Step1: Convert temperatures to Kelvin

For Charles's Law, temperatures must be in Kelvin.
$T_1 = 11 + 273=284\ K$
$T_2 = 121+ 273 = 394\ K$

Step2: Apply Charles's Law ($\frac{V_1}{T_1}=\frac{V_2}{T_2}$)

We know $V_1 = 1.33\times10^{6}\ L$, $T_1 = 284\ K$, $T_2 = 394\ K$. Rearranging for $V_2$ gives $V_2=\frac{V_1T_2}{T_1}$.
Substitute the values: $V_2=\frac{1.33\times 10^{6}\times394}{284}$
First, calculate $1.33\times394 = 523.02$
Then, $\frac{523.02\times10^{6}}{284}\approx1.84\times 10^{6}\ L$

Answer:

The volume of the air in the balloon after heating is approximately $1.84\times 10^{6}\ L$.